## Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage

# Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 358: 6

#### Answer

The areas of the four resulting smaller triangles is one fourth of the area of the given rhombus.

#### Work Step by Step

Draw diagonals MP and QN. the intersection of diagonals is named as O. Let us compare $\triangle$ OPN and $\triangle$ OQP we know in rhombus diagonals are perpendicular bisectors to each other So OP = OM and OQ = ON $\angle$ PON = $\angle$ POQ = $\angle$ MON = $\angle$ MOQ = 90$^{\circ}$ In $\triangle$ OPN and $\triangle$ OQP PN =QP(in rhombus opposite sides are equal) OP = OP (common) $\angle$ PON = $\angle$ POQ = 90$^{\circ}$ By SAS rule $\triangle$ OPN is congruent to $\triangle$ OQP Therefore area of $\triangle$ OPN = area of $\triangle$ OQP Since In rhombus sides are equal Similarly area of $\triangle$ OMN = area of $\triangle$ OMQ area of $\triangle$ PON = area of $\triangle$ OMN area of $\triangle$ OQP = area of $\triangle$ OQM so area of $\triangle$ OQP = area of $\triangle$ OPN = area of $\triangle$ OQM = area of $\triangle$ OQN Therefore area of MNPQ = area of $\triangle$ OQP + area of $\triangle$ OPN + area of $\triangle$ OQM + area of $\triangle$ OQN. Therefore area of MNPQ = 4 * area of $\triangle$ OQP area of $\triangle$ OQP = $\frac{area of MNPQ}{4}$. Therefore area of smaller triangle is one fourth of the area of rhombus MNPQ.

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